Problem 1 - Precalculus with Calculus Previews

Question

Use the laws of logarithms to show that for \(f(x)=\) \(\log _{b} x\) $$ \frac{f(x+h)-f(x)}{h}=\frac{1}{h} \log _{b}\left(1+\frac{h}{x}\right)=\frac{1}{x} \log _{b}\left(1+\frac{h}{x}\right)^{x / h}. $$

Step-by-Step Solution

Answer

The expression is simplified using logarithm laws, proving the equivalency as \(\frac{1}{h} \log_b \left( 1 + \frac{h}{x} \right) = \frac{1}{x} \log_b \left( 1 + \frac{h}{x} \right)^{x/h}\).

Step 1: Start with the Difference Quotient

The original expression is the difference quotient for the function \( f(x) = \log_b x \). It is given as:\[ \frac{f(x+h) - f(x)}{h} = \frac{1}{h} \log_b \left( \frac{x+h}{x} \right). \]

Step 2: Apply the Quotient Rule for Logarithms

Using the property of logarithms that \( \log_b \left( \frac{a}{b} \right) = \log_b a - \log_b b \), we rewrite the expression as:\[ \frac{1}{h} \left( \log_b(x+h) - \log_b x \right). \] This is equivalent to the initial difference \( f(x+h) - f(x) \).

Step 3: Simplify using the Property of Logarithms

Notice that \( \frac{x+h}{x} = 1 + \frac{h}{x} \), so the expression becomes:\[ \frac{1}{h} \log_b \left( 1 + \frac{h}{x} \right). \] This shows the equivalence in the original problem statement.

Step 4: Approaching the Given Expression

We need to show:\[ \frac{1}{x} \log_b \left( 1 + \frac{h}{x} \right)^{x/h}. \] Using properties of logarithms, specifically \( c \log_b a = \log_b a^c \):\[ \frac{1}{x} \log_b \left( 1 + \frac{h}{x} \right)^{x/h} = \frac{x/h}{x} \log_b \left( 1 + \frac{h}{x} \right) = \frac{1}{h} \log_b \left( 1 + \frac{h}{x} \right). \]

Step 5: Conclusion of Equality

Both sides of the expression have been transformed to \( \frac{1}{h} \log_b \left( 1 + \frac{h}{x} \right) \), thus proving the equivalency by using the laws of logarithms.

Key Concepts

Difference Quotient

The difference quotient is a crucial concept in calculus. It is used to approximate the tangent slope of a function at a particular point. For any function \( f(x) \), the difference quotient is given by:\[ \frac{f(x+h) - f(x)}{h} \]This expression represents the average rate of change of the function \( f(x) \) over the interval \( [x, x+h] \). The smaller \( h \) gets, the closer this average rate of change becomes to the instantaneous rate of change, which is the derivative.Understanding the difference quotient helps students transition from precalculus to calculus by introducing the concept of slopes of curves rather than simply lines.

Logarithmic Differentiation

Logarithmic differentiation is a technique used primarily for functions where direct application of differentiation rules becomes cumbersome. It relies on the natural properties of logarithms to simplify the differentiation process. Typically, it works best for functions that are products or quotients of terms raised to variable powers.Here's the basic approach:

• Take the natural logarithm of both sides of the equation \( y = f(x) \).
• Use logarithm properties to simplify.
• Differentiate using the chain rule.
• Solve for \( \frac{dy}{dx} \).

Logarithmic differentiation can be especially helpful in solving complex problems in precalculus and calculus.

Properties of Logarithms

Logarithms have unique properties that make them immensely useful in mathematics, especially for simplifying expressions and solving equations. Key properties include:

• Product Rule: \( \log_b(mn) = \log_b m + \log_b n \)
• Quotient Rule: \( \log_b\left(\frac{m}{n}\right) = \log_b m - \log_b n \)
• Power Rule: \( \log_b(m^k) = k \cdot \log_b m \)
• Change of Base Formula: \( \log_b a = \frac{\log_c a}{\log_c b} \)

These properties are essential in transforming and simplifying logarithmic expressions, as seen in our original problem where we use them to transition between different forms of the difference quotient.

Precalculus

Precalculus serves as a bridge between algebra and calculus. It equips students with a deeper understanding of mathematical concepts and prepares them for the rigorous logic of calculus. Key topics include:

• Functions and their properties
• Trigonometry and trigonometric functions
• Complex numbers
• Conic sections
• Vectors and polar coordinates

This foundational knowledge enables students to grasp more complex ideas such as limits, continuity, and differentiation—concepts critical for calculus. In our logarithmic problem, understanding precalculus principles aids in recognizing how logarithm properties apply within different mathematical contexts.